Monday, April 7th: “A” Day
Tuesday, April 8th: “B” Day
Agenda
Homework Questions/Collect
Go over Sec. 7.1 Quiz
Welcome Back!
Section 7.2: “Relative Atomic Mass/Chemical
Formulas”
In-Class Assignment:
Practice pg. 236: #1-2
Practice pg. 239-240: #1-4 (1 & 2: a,b,c,d only)
Homework: Concept Review: “Relative Atomic
Mass and Chemical Formulas”
Homework Questions/Problems?
Sec. 7.1 review, pg. 233…
Sec 7.1 Quiz:
“Avogadro’s Number and Molar
Conversions”
This quiz seemed to give some of you problems,
so I wanted to take some time to go over it…
Section 7.2: “Relative Atomic Mass
and Chemical Formulas”
You have learned to use the periodic table to
find the atomic mass of an element.
However, most atomic masses are written to
at least 3 decimal places.
Why?
Most Elements are a Mixture of Isotopes
 Remember: isotopes are atoms of the same
element that have different numbers of neutrons.
 Because they have different numbers of
neutrons, isotopes have different atomic masses.
 The periodic table reports average atomic mass,
which is a weighted average of the atomic mass
of an element’s isotopes.
 If you know the abundance of each isotope, you
can calculate the average atomic mass of an
element – and that’s what we’re going to do!
Rules for Calculating
Average Atomic Mass
1. Change each percentage to a decimal by
dividing by 100. (move the decimal point 2
places to the left and take away the % sign)
2. Multiply each decimal by the atomic mass
that goes with it.
3. Add the atomic masses together.
That’s it!
Calculating Average Atomic Mass
Sample Problem E, pg. 235
The mass of a Cu-63 atom is 62.94 amu, and that of
a Cu-65 atom is 64.93 amu.
(amu = atomic mass unit)
Using the data below, find the average atomic
mass of copper.
abundance of Cu-63 = 69.17%
 abundance of Cu-65 = 30.83%
Calculating Average Atomic Mass
Sample Problem E, continued
1. Change each percentage to a decimal (do not round):
 Cu-63 = 69.17% = .6917
 Cu-65 = 30.83% = .3083
2. Multiply each decimal by the atomic mass that goes with
it:
Cu-63 = (.6917) (62.94 amu) = 43.54 amu
Cu-65 = (.3083) (64.93 amu) = 20.01 amu
3. Add the atomic masses together:
43.54 amu + 20.01 amu = 63.55
amu
Additional Example #1
Chlorine exists as chlorine-35, which has a mass of 34.969
amu and makes up 75.80% of chlorine atoms. The rest of
naturally occurring chlorine is chlorine-37, with a mass of
36.996 amu. What is the average atomic mass of chlorine?
1. Change each percentage to a decimal:
 Cl-35 = 75.80% = .7580
 Cl-37 = 24.20% = .2420 (100% - 75.80%)
2. Multiply each decimal by the atomic mass that goes with
it:
Cl-35 = (.7580) (34.969 amu) = 26.51 amu
Cl-37 = (.2420) (36.996 amu) = 8.953amu
3. Add the atomic masses together:
26.51 amu + 8.953 amu =
35.46 amu
Additional Example #2
U-234 makes up 0.00500% of uranium atoms and has a mass
of 234.041 amu. U-235 makes up 0.720% and has a mass
of 235.044 amu. U-238 has a mass of 238.051 amu and
makes up 99.275%. What is the average atomic mass of
uranium?
1. Change each percentage to a decimal:
 U-234 = 0.00500% = 0.0000500
 U-235 = 0.720 % = 0.00720
U-238 = 99.275% = .99275
Additional Example #2, cont.
2. Multiply each decimal by the atomic mass that goes with
it:
U-234 = (0.0000500) (234.041amu) = 0.0117 amu
U-235 = (0.00720) (235.044) =
1.69 amu
U-238 = (.99275) (238.051) =
236.33 amu
3. Add the atomic masses together:
0.0117 amu + 1.69 amu + 236.33 amu =238.03
amu
Chemical Formulas and Moles
A compound’s chemical formula tells you which
elements, as well as how much of each, are present in
a compound.
Formulas for covalent compounds show the elements
and the number of atoms of each element in a
molecule.
Formulas for ionic compounds show the simplest
ratio of cations and anions in any pure sample.
Formulas Express Composition
Although any sample of compound has many atoms and
ions, the formula gives a ratio of those atoms or ions.
Formulas Give Ratios of Polyatomic Ions
Formulas for polyatomic ions show the
simplest ratio of cations and anions.
They also show the elements and the number
of atoms of each element in each ion.
For example, the formula KNO3 shows a ratio
of one K+ cation to one NO3- anion.
Calculating Molar Mass of Compounds
Sample Problem F, pg. 239
Find the molar mass of barium nitrate.
**Before you can find the molar mass, you need to write
the chemical formula**
 Barium is a 2+ cation and nitrate is a 1- anion.
Ba2+
NO3 Two NO3- anions are needed to balance the 2+ charge
on the barium cation.
 The simplest formula for barium nitrate is:
Ba(NO3)2
Calculating Molar Mass of Compounds
Sample Problem F, continued.
To find the molar mass of Ba(NO3)2, add the
atomic masses of each element together.
**The 2 outside of the ( ) means that everything
inside the ( ) is multiplied by 2**
Ba:
= 137.33 g/mol
N: 2 (14.01 g/mol) = 28.02 g/mol
O: 6 (16.00 g/mol) = 96.00 g/mol
Molar mass of Ba(NO3)2 = 261.35 g/mol
Additional Practice #1
Calculate the molar mass of ammonium sulfite,
(NH4)2SO3.
Add the atomic masses of each element
together:
N: 2 (14.01 g/mol) = 28.02 g/mol
H: 8 (1.01 g/mol)
= 8.08 g/mol
S: 32.07 g/mol
= 32.07 g/mol
O: 3 (16.00 g/mol) = 48.00 g/mol
Molar mass of (NH4)2SO3 = 116.17 g/mol
Additional Practice #2
Calculate the molar mass of aluminum sulfate,
Al2(SO4)3.
Add the atomic masses of each element together:
Al: 2 (26.98 g/mol)
= 53.96 g/mol
S: 3 (32.07 g/mol)
= 96.21 g/mol
O: 12 (16.00 g/mol)
= 192.00 g/mol
Molar mass of Al2(SO4)3
= 342.17 g/mol
In-Class Assignment/Homework
You Must SHOW WORK!
Practice pg. 236: #1,2
Practice pg. 239-240: # 1-4 (1 & 2: a,b,c,d only)
Homework:
Concept Review: “Relative Atomic Mass and
Chemical Formulas”
Next time:
Sec. 7.2 work day/quiz